Question

If sec a + tan a = p, then find the value of cosec a.​

Answer 1

$\huge\star{\green{\underline{\mathfrak{Answer :-}}}}$

Given :

• $\tt{ sec\:a\:-\:tan\:a =\:p}$.....(1)

To find :

• $\tt{cosec\:a}$

Solution :

We know that,

${\boxed{\tt{{sec}^{2}a-{tan}^{2}a =1}}}$

So,

$\leadsto\tt{(sec\:a\:-\:tan\:a)(sec\:a\:+\:tan\:a)=1}$

We know from the given that,

sec a - tan a = p.

Substituting it,

$\leadsto\tt{(p)(sec\:a\:+\:tan\:a)=1}$

$\leadsto\tt{(sec\:a\:+\:tan\:a)=\dfrac{1}{p}}$....(2)

Using elimination method in equation 1 and 2.

$\leadsto\tt{(sec\:a\:+\:tan\:a)=\dfrac{1}{p}}$

$\leadsto\tt{ sec\:a\:-\:tan\:a =\:p}$

$\rule{100}2$

$\leadsto\tt{2\:sec\:a = \dfrac{1}{p}+p}$

$\leadsto\tt{sec\:a=\dfrac{{p}^{2} +1}{2p}}$

$\leadsto\tt{\dfrac{1}{cos\:a} = \dfrac{{p}^{2} +1}{2p}}$

$\leadsto\tt{cos\:a=\dfrac{2p}{{p}^{2}+1}}$

$\leadsto\tt{sin\:a=\sqrt{1-{(\dfrac{2p}{{p}^{2}+1})}^{2}}}$

$\leadsto\tt{sin\:a=\sqrt{1-\dfrac{4{p}^{2}}{{({p}^{2}+1)}^{2}}}}$

$\leadsto\tt{sin\:a=\sqrt{\dfrac{{({p}^{2}+1)}^{2}-4{p}^{2}}{{({p}^{2}+1)}^{2}}}}$

$\leadsto\tt{sin\:a=\sqrt{\dfrac{{p}^{4}+12{p}^{2}-4{p}^{2}}{{({p}^{2}+1)}^{2}}}}$

$\leadsto\tt{sin\:a=\sqrt{\dfrac{{({p}^{2}-1)}^{2}}{{({p}^{2}+1)}^{2}}}}$

$\leadsto\tt{sin\:a=\dfrac{{p}^{2}-1}{{p}^{2}+1}}$

$\huge\leadsto{\boxed{\tt{\green{cosec\:a=\dfrac{{p}^{2}+1}{{p}^{2}-1}}}}}$

$\rule{200}2$

Answer 2

$\huge\sf{Answer:}$

Question:

If sec(a) + tan(a) = p.

Find:

= Value of cosec(a).

Solution:

= sec(a) - (a) = p.

Using formula;

= sec^2(a) - tan^(a) = 1.

= sec(a) - tan(a) (sec(a) + tan(a)) = 1.

= sec(a) - tan(a) = p.

Adding the above values:

= (p) sec(a) + tab(a) = 1

= (sec(a) + tan(a) = 1/p

Therefore, the value of cosec a = (sec(a) + tan(a) = 1/p.